However currently I don't know what does it mean. Further development in the book is rather obscure and I'm afraid of that "pseudo". If a thing called "pseudo-something" I would prefer it stated as "actual another thing".

UPDATE 2

Stress tensor can also be viewed as a (molecular) flux of momentum. Then the equation for balance of momentum would be the Newton's second law. Probably this approach would be more fruitful, analogues can be made with the flux of density.

My guess: $P$ is a $1$-form valued $2$-form. Surface force $f$ is also a $1$-form valued $2$-form, and power density is the $2$-form that results from contracting $f$ with the surface velocity.
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timurAug 8 '12 at 0:18

No, but as I said it is just a guess. I am curious why do you think it be star P?
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timurAug 8 '12 at 5:34

@timur I just don't know what is "1-form valued 2-form", I was guessing. Btw, see my update to the answer.
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YrogirgAug 8 '12 at 7:26

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A "foo valued" 2-form is, roughly speaking, something that, when combined with a bivector (or an ordered pair of tangent vectors), yields a "foo". The kind of n-form you're used to is a "scalar-valued" n-form.
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HurkylAug 8 '12 at 7:29

2 Answers
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I found a paper supporting my comment that $P$ is a 1-form valued 2-form, that surface force f is also a 1-form valued 2-form, and power density is the 2-form that results from contracting f with the surface velocity. The paper is

Given an inviscid fluid with a 0-form $p$ for preassure, how would you make a stress tensor for it?
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YrogirgAug 21 '12 at 12:47

@Yrogirg: Pressure is a 3-form (If you have a 0-form then just take its Hodge dual). The stress tensor $s$ corresponding to the 3-form $p$ is the following: Given a vector field $X$, the contraction $s(X)$, which is supposed to be a 2-form, is given by $i_Xp$.
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timurAug 23 '12 at 1:39

It is important to distinguish between covariant and contravariant indices of a tensor. Differential forms are totally antisymmetric covariant tensor fields. So a 2-form has 2 covariant indices, and when you swap them, the sign changes. Contravariant indices are written as upper indices and covariant indices as lower indices. You can raise and lower indices by use of a metric. Now, the stress tensor has one covariant index and one contravariant index. When you lower the contravariant index, you get a symmetric tensor field, not a differential form. In local coordinates, you simply have a matrix associated to every point, say ${\bf P}(\vec x)$.

The easiest way to understand what the stress tensor does is to imagine the effect of infinitesimal deformations inside the body, described by a vector field, say $\vec v(\vec x)$. The actual displacement at $\vec x$ could be written as $\vec v(\vec x) dr$. Now, the Energy density released by this displacement is $dE = P^j_iv^i_{;j}\ dr$, or, if you take $\vec v(\vec x)$ as a velocity, $P^j_iv^i_{;j}$ will simply be the power density. The semicolon indicates the covariant derivative. You can compute it by taking local coordiantes such that at $\vec x$ the metric is the Euclidean metric and all derivates of the metric are zero. In such local coordinates, $P^j_iv^i_{;j} = {\rm tr}({\bf PJ}_{\vec v})$, with ${\bf J}_{\vec v}$ the Jacobi matrix of $\vec v$.

Edit: What I am saying is that you cannot use differential forms alone. They are special tensors, but you need more general tensors. The stress tensor is a vector-valued 1-form (which, in 3 dimensions, is equivalent to a vector-valued 2-form, by Hodge duality, which gives a little more weight to the surface interpretation you formulated above). A vector is a contravariant 1-tensor, a 1-form is a covariant 1-tensor. Using the metric, you can transform one into the other, so you could even write the stress tensor as a 1-form-valued 1-form (or $(n-1)$-form, in $n$ dimensions), but that seems not very physical to me.

You can go from one to another using the metric, but I think the ideal formulation should be (or at least should try to be) independent of metric. An example I have in mind is Maxwell's electrodynamics, where the electric and magnetic fields are naturally 1- and 2-forms, and the metric enters only through the laws.
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timurAug 9 '12 at 23:55

@timur: What we need at any rate is a connection. We can make the stress tensor independent of the metric by multiplying it with the volume form: $P^j_i\omega_{klm}$ – we get a (vector-valued 1-form)-valued 3-form. We cannot make it independent of the connection though.
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Hendrik LönngrenAug 10 '12 at 11:32

@Yrogirg: The covariant derivative is coordinate-free, I just gave a method to compute it. You could write ${\rm tr}({\bf P\nabla}v)$ instead for the power density, which does not depend on coordinates.
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Hendrik LönngrenAug 10 '12 at 11:38

and then what is $\mathbf P$ in a coordinate-free view? What is the space it belongs to?
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YrogirgAug 10 '12 at 11:54